Rectilinear and Polygonal p-piercing and p-center Problems. Micha Sharir y Emo Welzl z. 1 Introduction. piercing points.

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1 Rectilinear and Polygonal p-piercing and p-center Problems Micha Sharir y Emo Welzl z Abstract We consider the p-piercing problem, in which we are given a collection of regions, and wish to determine whether there exists a set of p points that intersects each of the given regions. We give linear or near-linear algorithms for small values of p in cases where the given regions are either axisparallel rectangles or convex c-oriented polygons in the plane (i.e., convex polygons with sides from a xed nite set of directions). We also investigate the planar rectilinear (and polygonal) p-center problem, in which we are given a set S of n points in the plane, and wish to nd p axis-parallel congruent squares (isothetic copies of some given convex polygon, respectively) of smallest possible size whose union covers S. We also study several generalizations of these problems. New results are a linear-time solution for the rectilinear 3-center problem (by showing that this problem can be formulated as an LP-type problem and by exhibiting a relation to Helly numbers). We give O(n log n)-time solutions for 4-piercing of translates of a square, as well as for the rectilinear 4-center problem; this is worst-case optimal. We give O(n polylog n)-time solutions for 4- and 5-piercing of axis-parallel rectangles, for more general rectilinear 4- center problems, and for rectilinear 5-center problems. 2- pierceability of a set of n convex c-oriented polygons can be decided in time O(c 2 n log n), and the 2-center problem for a convex c-gon can be solved in O(c 5 n log n) time. The rst solution is worst-case optimal when c is xed. Both authors acknowledge support by G.I.F. the German Israeli Foundation for Scientic Research and Development, and by a Max Planck Research Award. Work by Micha Sharir has also been supported by National Science Foundation Grants CCR and CCR , and by grants from the U.S.{Israeli Binational Science Foundation, and the Israel Science Fund administered by the Israeli Academy of Sciences. Emo Welzl has also been supported by a Leibniz Prize from the German Research Society, We 1265/5-1. Part of the work on this paper was done during the participation of Emo Welzl in the Special Semester on Combinatorial and Computational Geometry, organized by the Mathematical Research Institute of Tel Aviv University. y School of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978, Israel, and Courant Institute of Mathematical Sciences, New York University, New York, NY 10012, USA. z Institut fur Informatik, Freie Universitat Berlin, and Departement Informatik, ETH Zurich, CH-8092 Zurich, Switzerland. 1 Introduction The problems. Let R be a set of n regions in the plane, and let p be a positive integer. R is called p-pierceable if there exists a set of p piercing points which intersects every member of R. The p-piercing problem is to determine whether R is p-pierceable, and, if so, to produce a set of p piercing points. In the p-center problem we are given a set S of n points in the plane, some compact convex set C, and a positive integer p. The goal is to nd p isothetic copies of C of smallest possible scaling factor, whose union covers S. This problem arises in the area of facility location in operations research, and many variants of it have been studied [7, 8, 10, 11, 19, 23, 25, 28, 30, 31, 33]. If C is the unit ball of some norm k k C, then the p-center problem seeks a set P of p `facility points', so that the maximum k k C-distance from a point of S to its nearest facility point is minimized. A natural case is when C is a unit disk, so k k C is the euclidean distance (euclidean p-center problem). When C is a square, then k k C can be viewed as the `1- or `1-norm (depending on the orientation of the square), and we face the rectilinear p-center problem. A standard reduction from the p-center problem to the p-piercing problem goes as follows. Let C ~ denote the re- ection of C (with respect to some interior point, which we assume to be the origin). Dene R() = fs + C ~ j s 2 Sg, 0. Then we seek the smallest possible for which R() is p-pierceable. The piercing points (for the smallest ) are the locations of the desired facilities; they serve as centers of isothetic copies of C whose union covers S. Often, the p-center problem is solved by techniques like parametric searching [29] or monotone matrix searching [13, 14, 15], which run some sort of binary search on to locate the optimal solution. The related p-piercing problem, for any xed, is then referred to as the decision problem, or the xedsize problem, and is used as an `oracle' to guide the binary search. Previous results. It is known that the p-piercing and p-center problems are NP-complete when p is part of the input, even if the regions are translates of a square [31], and that they can be solved in polynomial time for any xed p (assuming that C and the sets to be pierced are simply shaped, see [7, 24]). The 1-piercing problem is easy for simple regions, and the 1-center problem is well understood at this point and allows optimal linear-time solutions for many types of regions we will not elaborate on this here (see [27, 34]). The euclidean 2- center problem was recently solved in near-linear time [33], but no such solution is known for p 3. The 2- and 3- piercing problems for a set of axis-parallel rectangles have

2 been solved in linear time [22, 23]. 1 The rectilinear 2-center problem was also known to be optimally solvable in linear time [8], while the best previous solution for the rectilinear 3-center problem takes O(n log n) time [8, 23]. Actually, the papers [22, 23] consider a slightly more general version, the so-called weighted rectilinear p- center problem: Here we are given n axis-parallel squares, not necessarily of the same size. We ask for the smallest common scaling factor, so that the scaled squares (where the scaling is about the center of each square) is still p- pierceable. It is shown in [22] that the weighted rectilinear 2-center problem is solvable in linear time. 2-pierceability of c-oriented convex polygons is closely related to a conjecture of Danzer and Grunbaum [6], which asks for a Helly-type theorem for collections of homothets of a convex polygon, and which was recently refuted in [20]. For p > 3, the best previous bound for the rectilinear p-center problem is O(n p 2 log n) [24]. No near-linear solutions have previously been known for the p-piercing or p-center problem in the general polygonal case, for p 2. New results. We improve the O(n log n) bound for the (weighted) rectilinear 3-center problem to optimal linear time. For the (unweighted) rectilinear 4-center problem we give an O(n log n) solution, and for the rectilinear 5-center problem we describe an O(n log 5 n) solution, thereby improving on the previous bounds of the form O(n 2 log n) and O(n 3 log n), respectively. The 4-center algorithm is also worst-case optimal. The new bounds for the center problems are based on new results for the corresponding piercing problems: 4-piercing of translates of a square can be decided in O(n log n) time, 4- (and 5-) piercing of general axis-parallel rectangles can be decided in time O(n log 3 n) (and O(n log 4 n), respectively). These results also lead to O(n polylog n) algorithms for the weighted rectilinear 4- and 5-center problems. Our solutions also imply that the rectilinear p-center problem, for any p 5, can be solved in time O(n p 4 log 5 n). We believe, however, that much better results can be obtained for larger p's. The 2-center problem, when C is a convex c-gon (polygonal 2-center problem), can be solved in O(c 5 n log n) time. This is based on 2-piercing of c-oriented convex polygons (whose sides draw their directions from a xed set of c orientations), for which we supply an O(c 2 n log n) solution, and show that it is worst-case optimal when c is xed. 2-piercing of a set of homothetic triangles, and the triangular 2-center problem, can both be solved in linear time. The same holds for 4-oriented trapezoids. Independently, Katz and Nielsen [21] have recently obtained several results on the piercing problems studied in this paper, and also results concerning 2-piercing of boxes, simplices, and c-oriented polytopes in higher dimensions. They have not addressed the related center problems. Methods. In order to obtain the linear bound for the 3-center problem, we formulate it as an LP-type problem 1 In fact, the results in [23] are stated only for homothets of squares. [26], from which a randomized linear-time algorithm follows, which can also be derandomized to a deterministic linear-time solution [5]. The proof of the LP-type properties (and of nite combinatorial dimension see below) relies on a Helly-type result of the following kind [6]: A set of axisparallel rectangles is 3-pierceable if and only if any 16 (or fewer) of them are 3-pierceable. The connection between Helly-numbers and LP-type problems has been investigated in [3, 4]. The realization of the rectilinear 3-center problem as an LP-type problem seems to be particularly interesting, since there is no formulation of this problem as a convex optimization in a convex domain, which shows that LP-type problems go beyond convex programming (in natural examples). The basis for the rectilinear piercing results are a number of simple observations on the location of piercing points, which have already been used (in one or another disguise) in previous attacks on the rectilinear problem [8, 22, 23]. We enhance those insights by a carefully planned iterative process (for 4-piercing of translates of a square), and by multi-level data structures (for general rectilinear 4- and 5- piercing). These observations can also be extended (but become slightly more involved) to the case of c-oriented polygons, which leads to an O(n log n) solution of the 2-piercing problem for such polygons. (The case of triangles or of trapezoids can be formulated as an LP-type problem, but there is also a simple deterministic linear-time solution.) For both the standard rectilinear and polygonal cases, the reduction from the center to the piercing problem uses the matrix-searching technique of Frederickson and Johnson [13, 14, 15], in a manner similar to recent applications in [17]. For the weighted rectilinear and polygonal cases, we employ standard parametric searching techniques [29]. 2 Rectilinear p-piercing We are given a set R of n axis-parallel closed rectangles in the plane and some positive integer p, and we wish to decide whether R is p-pierceable. We start with some basic observations for general p (see also [8, 22, 23]), and then \specialize" to p = 2; : : : ; 5 to obtain linear or near-linear solutions. An axis-parallel line traverses R. Suppose a vertical line ` intersects all rectangles in R (the case where ` is horizontal is treated similarly), and let P be some set of piercing points. Consider a 2 P. If we move a on a horizontal segment towards `, then we do not leave any of the rectangles (we may enter some). That is, if we replace each a 2 P by its horizontal projection a 0 on `, then the resulting set P 0 is still piercing. Consequently, the p-piercing problem can be solved by deciding p-pierceability of the intervals R 0 = f` \ r j r 2 Rg on `. This 1-dimensional piercing problem can be easily solved in linear time (see, e.g., [18, p. 193]). So if there is a vertical (or horizontal) line intersecting all rectangles, the problem can be solved in linear time for any xed p. No axis-parallel line traverses R. Let `L be the vertical line containing the leftmost right edge of a rectangle in 2

3 `L R 0 Figure 1: A 4-pierceable set of axis-parallel rectangles and its location domain R 0 R, let `R be the vertical line containing the rightmost left edge of a rectangle in R, let `T be the horizontal line containing the highest bottom edge of a rectangle in R, and let `B be the horizontal line containing the lowest top edge of a rectangle in R. Consider the closed left halfplane H L bounded by `L, the closed right halfplane H R bounded by `R, the closed top halfplane H T bounded by `T, and the closed bottom halfplane H B bounded by `B. Let H ~ X denote the closure of the complement of H X, for X = L; R; T; B. Note that if no axis-parallel line traverses R, then H L is disjoint T from H R, and H T is disjoint from H B, and thus R 0 := ~ X2fL;R;T;Bg H X is nonempty (with nonempty interior). R 0 is called location domain for R, a term justied in the following simple but crucial observation. Observation 2.1 (a) Any set of piercing points of R must have a point in each of the halfplanes H X, for X = L; R; T; B. (b) Assume that R is p-pierceable, but has no axis-parallel traversing line. Then there is a set of p piercing points in the location domain R 0. In such a piercing set, each side of the 0 of R 0 must contain one of the piercing points. (Here we consider the sides as relatively closed, that is, a vertex 0 is contained in its two incident sides.) See Figure 1. (a) is trivially true, since each halfplane H X fully contains one of the rectangles in R, for X = L; R; T; B, and so it must contain a piercing point. For (b), consider a piercing set P and a point a 2 P which is not in ~ H L, i.e., it is to the left of `L. Similar to a previous argument, we consider the horizontal projection a 0 of a on `L. The set of rectangles containing a 0 is a superset of the set of rectangles containg a. Hence we can replace a by a 0 in P without losing its piercing property, and a 0 is contained in ~ H L. A similar procedure can be employed if a piercing point is not in ~ H R, ~H T, or ~ H B. In this way we can move every piercing point that lies outside the location domain R 0 to its boundary (up to two steps may be necessary for each point). After that, the piercing set is contained in R 0, but (a) is still valid, and so each side of R 0 must contain a piercing point and 3-piercing If we want to decide whether a set of axis-parallel rectangles is 2-pierceable, we rst determine `L and `R. If `L is not to `R `T `B the left of `R, then a vertical line (say `L or `R) intersects all rectangles and we can invoke the simple 1-dimensional decision procedure. Otherwise, we apply a symmetric procedure for `T and `B. What remains is the case that the location domain R 0 is a nonempty rectangle. Observation 2.1(b) states that if R is 2-pierceable, then it is pierceable by two points in R 0, with one point on each side of R 0, and the only candidates for such a set are the two diagonal pairs of vertices 0. We can check these two possibilities in linear time. For 3-pierceability, we can either apply the 1-dimensional procedure, or R 0 exists, and now, by Observation 2.1(b), a piercing set of 3 points contained in R 0 must contain one of the vertices 0. We can take each of those vertices v, and check for 2-pierceability of the rectangles disjoint from v. If one of the four possibilities is successful, then R is 3-pierceable; otherwise, it is not. Summing up, we get the following result, which is at least implicitly also implied by [8, 22, 23]. Theorem and 3-pierceability of a set of n axisparallel rectangles can be decided in O(n) time. Remark. One can make the above algorithms more concise to state and implement by writing down an explicit list of a constant number of candidate sets of pairs or triples of piercing points, which depend on the given set of rectangles, so that this set is 2- or 3-pierceable if and only if one of these sets is piercing. We will do this below, when we talk about the corresponding 2- and 3-center problems piercing of translates of a square Let Q be a set of n translates of a square of unit side length, for which we want to decide the existence of 4 piercing points. Again, we may assume that the location domain R 0 exists. We can easily decide whether there is a piercing set of four points, one of which is a vertex 0: Take each such vertex in turn, and test for 3-pierceability of the squares disjoint from it, in linear time. So if all these attempts to nd a set of four piercing points fail, and Q is 4-pierceable, then we know, by Observation 2.1(b), that there must be four piercing points, one in the relative interior of each of the sides of R 0. Denote those sides (relatively open segments) by s X, for X = L; R; T; B. We manipulate a set Q 0 (initially Q 0 = Q), and four segments t X, for X = L; R; T; B (initially t X = s X ), while maintaining the following invariant: Q is pierceable by four points, one in each s X if and only if Q 0 is pierceable by four points, one in each t X. The invariant is trivially true at the initialization stage. The operations we perform are the following: (i) If a square q in Q 0 intersects only one of the segments t X, then we replace 2 t X by t X \ q and remove q from Q 0. (ii) If a square q contains one of the segments t X, then we remove q from Q 0. Clearly, the invariant is maintained after each operation of type (i) or (ii). Note that we can apply right at the start four operations of type (i), to the four squares dening R 0 2 Note that, as soon as a segment t X is shortened, it is not relatively open any more. 3

4 (as in Figure 1). We may end up in one of the following three situations: (A) At some point a square in Q 0 is disjoint from all segments t X. That is, Q 0 is not pierceable by points on the segments t X, and the invariant implies that Q is not pierceable by points, one on each s X. With the previous assumptions, we conclude that Q is not 4-pierceable. (B) At some point Q 0 is empty. We conclude that Q is 4- pierceable. (Note here that the segments t X can never vanish in the process.) (C) Neither (i) nor (ii) can be applied, Q 0 is still nonempty, and no square in Q 0 is disjoint from all segments t X. That is, each square intersects at least two of the segments t X, and no square contains one of those segments. Actually, both conditions combined imply that every square intersects exactly two segments (because, if a square intersects three of the sides of R 0, it must contain one of the sides). Note that in all we have said so far we never used the fact that we are handling congruent squares; everything also holds for general axis-parallel rectangles. We now claim that, for translates of a square, case (C) already implies 4- pierceability. For that purpose, let P 0 be the set of counterclockwise endpoints of the segments t X (the rst endpoint of each segment that we encounter as we 0 in clockwise direction), and let P 00 be the set of clockwise endpoints. We want to prove that either P 0 or P 00 is a piercing set for Q 0. Any square in Q 0 containing the upper right vertex of R 0 must intersect t T and t R ; more precisely, it must contain the counterclockwise endpoint of t R and the clockwise endpoint of t T. In other words, such a square is pierced by both P 0 and P 00. More generally speaking, a square containing a vertex of the location domain is pierced by both P 0 and P 00. Any square q in Q 0 disjoint from the vertices of R 0 must intersect two opposite sides of R 0 in order to meet two of the segments t X. Note that if q intersects the top and bottom side of R 0, but is disjoint from the vertices, then those sides must have length exceeding one and so no square in Q 0 can intersect both the left and right side of R 0 (because all squares are translates of a unit square). So, without loss of generality, we may assume that all squares disjoint from the vertices intersect the top and bottom sides of R 0. The segments t X have length at most one. Hence it is not possible for a square in Q 0 to intersect t T (and t B ) without containing one of its endpoints. If a square contains the counterclockwise endpoint of one of the two segments, and the clockwise endpoint of the other segment, then it is pierced by both P 0 and P 00. What is left are squares that only contain the clockwise endpoints of t T and t B, and squares that only contain the counterclockwise endpoints of t T and t B. If only one of the two situations occurs, then P 00 (or P 0, respectively) pierces all squares. Simultaneous occurrence of both situations, however, is impossible: A square q of the rst type has the counterclockwise endpoint of t T to its left, and the counterclockwise endpoint of t B to its right, and no square of unit side length can contain these two points. We have shown that either P 0 or P 00 is a piercing set in situation (C). For the time analysis, it suces to consider the algorithm for deciding 4-pierceability for the case that no axis-parallel line intersects all squares, and there is no piercing set containing a vertex of R 0 { those decisions can be made in linear time, and handling either of these situations can be done in linear time, as noted earlier. Now we simply have to perform operations (i) and (ii) until we end up in one of the situations (A), (B), or (C), when we can decide 4-pierceability in constant time (and produce a piercing set in linear time). In order to execute the operations (i) and (ii) eciently, we sort the intersections of the boundaries of the squares in Q with the sides of R 0 along each such side (there are at most 4n such intersections). Now, whenever we shorten a segment t X, we have to detect those squares which lose contact with t X, or start to contain t X. Any such square must have a side that intersects the part removed from t X. We therefore simply traverse this part, and process each intersection point found there in a straightforward manner. The most expensive part of the procedure is the sorting along the sides of R 0; all other steps can be performed in O(n) time. Note that if the centers of the squares have already been sorted both by their x-coordinates and by their y-coordinates, then the sorting along the sides of R 0 can be done in O(n) time, by merging the (already sorted) sequences of left sides and of right sides (or of top sides and of bottom sides) of the squares. We will use this observation in the 4-center algorithm, given below. We thus have: Theorem 2.3 (a) The 4-piercing problem for n translates of a square can be solved in O(n log n) time. (b) If the centers of the squares are presorted by their x- coordinates and by their y-coordinates, then 4-pierceability can be decided in O(n) time piercing of rectangles We are given a collection R = fr 1; : : : ; r ng of n axis-parallel rectangles in the plane, and we wish to determine whether R can be pierced by four points. Again, as in the preceding subsection, we may restrict ourselves to the situation that all four piercing points lie on the boundary of the location domain, each lying in the relative interior of a distinct side. This case is handled as follows. Let s denote the top side of R 0, and let t denote the intersection of s with the rectangle whose bottom side lies on the line `T. Let I j = t \ r j, for j = 1; : : : ; n. The endpoints of the intervals I j partition t into at most 2n 1 `atomic' intervals. We iterate through these intervals from left to right, and attempt to place the top piercing point p T in each of them. For an atomic interval J, the set R J of rectangles of R that do not contain p T remains unchanged as p T varies within J. When we pass from J to the next interval J 0, the set R J 0 diers from R J by a single rectangle being added or removed. For each interval J we check whether R J is 3-pierceable. If this is true for at least one interval J, then R is 4-pierceable; otherwise R is not 4-pierceable. We next describe a dynamic data structure on the rectangles in R J, using which we can determine whether R J is 3-pierceable in O(log 3 n) time; the structure can also be updated in O(log 3 n) time when a rectangle is added to or is removed from R J. Recalling the algorithm for testing for 4

5 3-pierceability, we see that the data structure has to support an ecient implementation of two main operations: (a) For a given pre-stored (`canonical') subset of rectangles of R, nd the corresponding lines `L, `R, `T, `B. (b) For a given pre-stored subset R 0 of rectangles of R and for a query point v, nd the set R 0 v of rectangles of R 0 that do not contain v. In a single step of the algorithm, both operations will be performed on a polylogarithmic number of canonical subsets, organized in a three-level structure (described in detail shortly), whose union is equal to a specic subset R. It is important to observe that both operations (a) and (b) are decomposable, in the sense that the output for R can be easily obtained from the outputs for the canonical subsets of R, in time proportional to the number of subsets: For (a), one simply has to take the leftmost of all the lines `L produced for each canonical subset, and symmetrically for the other three lines. The output of (b) for R is simply the union of the outputs for the canonical subsets of R. To perform both (a) and (b) on a canonical set R 0, we prepare four balanced binary search trees, storing, respectively, the left, right, top, and bottom edges of the rectangles of R 0, sorted by their x-coordinates (for the rst two trees) or by their y-coordinates (for the last two trees). Each node u of each of these trees represents a canonical subset R 0 u of R 0, consisting of all rectangles stored at the leaves of the subtree rooted at u. Unless we are at the bottommost (third) level of the structure, we construct, in the next deeper level of the structure, a similar structure for each of the sets R 0 u. Performing an operation of type (a) is easy: simply pick up the rst or last leaf of each tree, as appropriate, and use the corresponding rectangle sides to construct the desired lines. Performing an operation of type (b) is also easy. Suppose the query point v has coordinates (v x; v y). Search with v x in the rst two trees, and with v y in the last two trees. In the rst tree, obtain all rectangles whose left side lies in the halfplane x > v x, as the disjoint union of O(log n) subtrees. Apply symmetric procedures to the three other trees. In total, we obtain the output to (b) as the (not necessarily disjoint) union of O(log n) canonical subsets (that is, subtrees of the four trees). In order to facilitate simple and ecient updating of the structure, we use a slightly modied variant, in the spirit of the structure in [32]. We give details in the full version, and show that each insertion or deletion can be performed in O(log 3 n) time. We can now describe the full algorithm: (i) Construct the rectangle R 0 and obtain the partition of the top interval t, as dened above, into up to 2n 1 atomic intervals. (ii) Iterate over these intervals from left to right. For each interval J, maintain the set R J of rectangles of R not containing J. Initialize the data structure with the set of rectangles not containing the left endpoint of t. (iii) Let R 0(J) denote the location domain for R J. It is easily seen that the lines incident to the left, right and bottom sides of R 0(J) are the same lines `L, `R, `B dening R 0, and only the top side of R 0(J) changes (assuming general position, it is now lower than `T ). Find the new top side (an operation of type (a)). (iv) It is easily seen from our assumptions on the location of the piercing points that one of the three piercing points of R J, if they exist, can be assumed to be one of the two top vertices of R 0(J). We attempt to place the second piercing point at each of these vertices. Let v be that vertex, and, without loss of generality, assume it is the top-left vertex of R 0(J). (v) We compute the set R (J;v) of all rectangles in R J that do not contain v; this is an operation of type (b), and its output consists of O(log n) canonical subsets, whose union is R (J;v). We need to test whether R J;v is 2-pierceable. Again, our assumptions on the piercing points are easily seen to imply that the two piercing points can be assumed to be the top-right vertex and the bottom-left vertex of the corresponding location domain R 0(J; v). It is also easily veried that the bottom and right edges of R 0(J; v) lie on the same lines `B, `R as the corresponding sides of the original R 0. (vi) We nd the left side of R 0(J; v). This is done by computing the leftmost right edge of each of the canonical subsets produced in step (v) above, accessing the corresponding trees in the second level of our data structure, and by nding the leftmost of those output edges. In the same manner we nd the top side of R 0(J; v). The total cost of this step is O(log 2 n). Let w and z denote, respectively, the top-right vertex and the bottom-left vertex of R 0(J; v). (vii) We next determine whether R (J;v) is 2-pierceable by the points w and z. This is done in a manner similar to that used in the preceding steps, but by accessing the third level of the structure; we omit the description due to lack of space. The total cost of this step is easily seen to be O(log 3 n). (viii) If R J has been determined not to be 3-pierceable, we move to the next atomic interval J 0 of t. We update our data structure with the insertion or deletion of the rectangle by which R J and R J 0 dier, and repeat the above steps to J 0. As mentioned above, the cost of the update is O(log 3 n). (ix) If none of the R J's is found to be 3-pierceable, we conclude that R is not 4-pierceable. The overall running time of the algorithm is easily seen to be O(n log 3 n). Hence we have: Theorem 2.4 The 4-piercing problem for n axis-parallel rectangles can be solved in O(n log 3 n) time piercing Let R be a collection of n axis-parallel rectangles, as above. We want to determine whether R can be pierced by ve points. Let R 0 denote the location domain of R as dened above. Arguing as in the preceding subsections, we may assume that one of the following situations occurs: (i) One of the piercing points lies at a vertex of R 0. (ii) The ve piercing points all lie on the boundary of R 0, but none of them lies at a vertex. Each side of R 0 contains at least one point (and one side contains two points). (iii) Four of the piercing points lie on the boundary of R 0, with one point lying in the relative interior of each side, and the fth point lies in the interior of R 0. 5

6 (i) (ii) (iii) Figure 2: Guessing the bottom piercing point (in cases (i,ii)) or the top piercing point (in cases (ii,iii)) causes one of the 4 remaining piercing points to lie at a vertex of the location domain R 0 (J). Testing for case (i) is easy: We try each of the four vertices of R 0 as the rst piercing point, nd the set of rectangles not containing that vertex, and test whether this set is 4-pierceable, using the preceding algorithm. This takes O(n log 3 n) time. To test for case (ii), we guess a side s of R 0 that contains just one piercing point, say it is the top side, and apply a procedure similar to that given for the 4-piercing problem. That is, we construct the interval t and its partition into atomic subintervals, as above, and iterate over these subintervals from left to right, maintaining the set R J of rectangles not containing the current interval J. It is easy to check that one of the four remaining piercing points must lie at one of the two top vertices of R 0(J), the location domain for the set R J. We compute R 0(J) and attempt to place the second piercing point at either of its top vertices, call it v. We nd the set R (J;v) of those rectangles of R J that do not contain v, and test whether this set is 3-pierceable, following the procedure described in the preceding section. The data structure that we need here is almost identical to that used above, except that now it has four levels instead of three. Omitting the easy missing details, we obtain a procedure that runs in O(n log 4 n) time. To test for case (iii), let q be the piercing point lying inside R 0. Either at least one of the piercing points on the left and right sides of R 0 lies above q or at least one of these points lies below q. See Figure 2. Suppose that one of them lies above q. We then `guess' the piercing point lying on the top side of R 0, using the interval t and its partition into atomic subintervals, as above. For each atomic interval J, we maintain the set R J of rectangles not containing J, and observe that, in the assumed conguration, one of the four piercing points of R J, if they exist, must be a (top) vertex of R 0(J). This is because R 0(J) diers from R 0 only by its top side, which is lower, and must pass through one of the two piercing points lying on the left and right sides of R 0, because the highest of these two points lies above q, by assumption. It therefore suces to test whether R J is 4-pierceable, with one piercing point lying at a vertex of R 0(J). This can be tested as in case (ii) above, implying that the testing for case (iii) can also be done in O(n log 4 n) time. Hence we obtain: Theorem 2.5 The 5-piercing problem for a set of n axisparallel rectangles can be solved in O(n log 4 n) time. Remark. The above approach fails for the 6-piercing problem. One reason is that two of the piercing points might lie inside R 0 in such a way that, no matter which of the four border piercing points we eliminate, the resulting 5-piercing subproblem need not have a piercing point at a vertex of the associated location domain. In this case we do not know how to test for this 5-pierceability in polylogarithmic time. The best algorithm for 6-pierceability that we can design takes near quadratic time (it guesses simultaneously two piercing points on the boundary of R 0). We do not know whether this problem can be solved in subquadratic time. Remark. Extending the last observation, one can easily show that the general rectilinear p-piercing problem, for p 5, can be solved in time O(n p 4 log 5 n). 3 Rectilinear p-center Problems We choose the following set-up of the problem: Let R be a set of n compact convex regions with nonempty interior, where every region r 2 R is assigned a scaling point c r in its interior. For r 2 R and a real number 0, let r() be the homothetic copy of r obtained by scaling r by the factor about c r (i.e. r() = fc r + (a c r) j a 2 rg). Finally, R() = fr() j r 2 Rg. The p-center problem for R asks for R := minf j R() is p pierceableg : If R is a set of translates of a square and the scaling points are the respective centers, then we talk about the (standard) rectilinear p-center problem. If the squares are still axisparallel but of possibly dierent sizes (and again the scaling points are the centers), then we have the weighted rectilinear p-center problem, and if R is a set of arbitrary axis-parallel rectangles (and the scaling points are also arbitrary), then we face the general rectilinear p-center problem. 3.1 General rectilinear 2- and 3-center problems Given a set R of axis-parallel rectangles with scaling points, we want to determine 00 R (and 000 R), the smallest for which R() is 2-pierceable (3-pierceable, respectively). We will use two tools for establishing a linear time bound, the LP-type framework and Helly-type results, which we review rst. LP-type problems. An LP-type problem is a pair (H; w) where H is a nite set (whose elements are called constraints), and w is a mapping from 2 H into some totally ordered set, so that the following conditions are satised: (Monotonicity) For any F; G with F G H, we have w(f ) w(g). (Locality) For any F G H with w(f ) = w(g) and any h 2 H, w(g) < w(g [ fhg) implies that also w(f ) < w(f [ fhg). A basis B is a set of constraints with w(b 0 ) < w(b) for all proper subsets B 0 of B. B is a basis of G, for G H, if B G is a basis and w(b) = w(g). 6

7 LP-type problems serve as a framework to solve certain optimization problems. Solving an LP-type problem (H; w) means to determine a basis of H, based on the following basic operations. (Violation test) For a constraint h and a basis B, test whether h is violated by B, i.e. whether w(b) < w(b [ fhg). (Basis computation) For a constraint h and a basis B, compute a basis of B [ fhg. For the eciency of algorithms solving LP-type problems, the following parameter is crucial: The maximum cardinality of any basis is called the combinatorial dimension of (H; w). Lemma 3.1 ([26]) An LP-type problem of combinatorial dimension d with n > d constraints can be solved by a randomized algorithm with an expected number of at most 2 d+3 (n d) basic operations. Helly-type results for rectilinear p-piercing. For p 1, let h p;d be the smallest integer such that a nite set of axis-parallel boxes in d dimensions is p-pierceable, if and only if any h p;d (or fewer) of the boxes are p-pierceable. Lemma 3.2 [6] (cf [9], Theorem 5.5) (a) h p;1 = p + 1, for any p 1. (b) h 1;2 = 2, h 2;2 = 5, h 3;2 = 16, and h p;2 is undened for p 4. (c) For d 3, one has h 1;d = 2, h 2;d = 3d if d is even and h 2;d = 3d 1 if d is odd, and h p;d is undened for p 3. For our purpose, this yields the following corollary (whose easy proof is omitted in this abstract). Corollary 3.3 If R is a set of axis-parallel rectangles with scaling points, then there exists a subset G 2 of at most 5 rectangles such that 00 R = 00 G 2 and there is a subset G 3 of at most 16 rectangles such that 000 R = 000 G 3. The relation between nite Helly numbers and LP-type problems of constant combinatorial dimension has been worked out in [3, 4]. Roughly speaking, as shown there, constant combinatorial dimension implies a nite Helly number property, but not vice versa! The p-center problem in one dimension as an LPtype problem. Before considering the 2-dimensional case, let us comment on the 1-dimensional p-center problem. A fairly involved technique of Frederickson [12] gives a linear-time solution, provided that the points are already sorted along the real line (this result actually works for arbitrary trees). However, it follows from Lemma 3.2(a), combined with the machinery of [3, 4], that the 1-dimensional p-center problem is an LP-type problem, for any p (see below for a similar analysis in two dimensions), and thus can be solved in randomized expected O(n) time, for any xed p. In this bound, though, the combinatorial dimension is O(p), so the constant of proportionality depends `subexponentially' on p (see [26]), so Frederickson's solution is far better when p is large. The 2-center problem as an LP-type problem. (Recall that the 2-center problem already has a linear-time solution [8]. It is treated here both to highlight the use of the LP-type machinery to solve this problem, and to serve as a basis for the solution of the 3-center problem.) If we simply dene w(g) := 00 G for G R, then the system (R; w) is not LP-type because locality may be violated (this is discussed in more detail in the full version). We need to dene a unique canonical 2-piercing set of a set of rectangles, provided it is 2-pierceable, and use this canonical set as part of the value of w. Recall the denition, given in the beginning of Section 2, of the lines `X, for X = L; R; T; B, for a nonempty set R of axis-parallel rectangles. Dene the vertex v XY as the intersection of `X and `Y for X = L; R and Y = B; T. Given a set R of axis-parallel rectangles, we dene six 2-piercing candidate sets by C (0) = ;; C (1) = fv RT g; C (2) = fv LT ; v RT g; C (3) = fv RB ; v RT g; C (4) = fv LB ; v RT g; C (5) = fv LT ; v RB g : Suppose that R is 2-pierceable. If R is empty then C (0) pierces R. If R is 1-pierceable, then C (1) pierces R. If there is a horizontal line intersecting all rectangles, then C (2) pierces R, and if a vertical line intersects all rectangles, then C (3) pierces R. If no axis-parallel line intersects all rectangles, then the location domain exists, and C (4) or C (5) is piercing. We summarize in a lemma. Lemma 3.4 If R is 2-pierceable, then one of the 2-piercing candidate sets pierces R. The canonical 2-piercing set P R of R is the 2-piercing candidate of smallest index that pierces R, and the index of this candidate is called the characteristic 2-piercing index R of R. If R is not 2-pierceable, then we set R = 6 and P R =?, with? representing a nominal undened value. Now let R be a set of axis-parallel rectangles with scaling points assigned. For G R, we dene w(g) as the tuple ( 00 G; x L ; x R ; y B ; y T ; G( 00 ), where G ) x L, x R are the x-coordinates of `L, `R, and y B, y T are the y-coordinates of `B, `T, as dened for the scaled set G( 00 G). Note that w(g) describes a unique candidate piercing set, including the exact coordinates of the piercing point(s). The ordering on such tuples is lexicographic, with the rst entry most signicant. We have to show monotonicity of w. As for the rst component, it is clear that 00 F 00 G for F G R. The following lemma claries that adding a rectangle to a set G cannot decrease its characteristic 2-piercing index. Lemma 3.5 Let R be a 2-pierceable set of axis-parallel rectangles. (0) R = 0 i R is empty. (1) R = 1 i R is nonempty and 1-pierceable. (2) R = 2 i R is not 1-pierceable, and there is a horizontal line intersecting all rectangles. (3) R = 3 i R is not 1-pierceable, and there is a vertical line intersecting all rectangles. (4) R = 4 i there is no axis-parallel line intersecting all rectangles, and C (4) is piercing. (5) R = 5 i there is no axis-parallel line intersecting all rectangles, and C (4) is not piercing. 7

8 The conditions for (2) and (3) are disjoint because, if there is both a vertical and a horizontal line intersecting all rectangles, then the set is 1-pierceable. A short scan through the items in the lemma shows that adding a rectangle cannot decrease the characteristic 2- piercing index, with the step from (5) to (4) being the only non-trivial case. To see also this case, assume that we add a rectangle to R with R = 5, and the resulting set ^R is (4) pierceable by its fourth candidate ^C = f^v LB ; ^v RT g. Compare the points with the fourth candidate C (4) = fv LB ; v RT g of R. ^v LB must lie in the third quadrant anchored at v LB, and thus moving from ^v LB to v LB we cannot exit a rectangle in R. Similarly, v RT pierces all rectangles in R which are pierced by ^v RT (4). So if ^C is piercing ^R (and thus R), then C (4) is piercing R, contradicting the assumption that R = 5. So the rst component in the w-value of a set cannot decrease if a new rectangle is added. The same holds for all the other components, as is easily veried for the next four components, and as follows from the lemma for the last component. We have thus established monotonicity of w. As for locality, consider F G R with w(f) = w(g) = (; x 1; x 2; y 1; y 2; ) and a rectangle r with a scaling point. Note that w(g) < w(g [ frg), i r() is disjoint from the piercing candidate set described by w(g), which is equivalent to the condition w(f) < w(f [ frg). This proves locality. We are left with the task of proving that the combinatorial dimension (R; w) is bounded by a xed constant. That is, we need to bound the number of rectangles in a set G R that determine w(g). At most ve rectangles determine 00 (G). At most four other rectangles may be needed to x the lines `X, for X = L; R; T; B (and thus x the points v XY, for X = L; R and Y = B; T.) If G = i, then we may need up to i 1 additional rectangles to enforce that index (for each 1 j < i we need a rectangle r 2 G, so that r( 00 G) is disjoint from C (j) ). We obtain an upper bound of = 13 on the maximal size of a basis (remember that G( 00 G ) < 6). The actual bound is probably much smaller, since we can `recycle' some of the rectangles in the argument. The two types of basic operations are easy to implement in constant time. Lemma 3.1 implies: Theorem 3.6 The general rectilinear 2-center problem for n rectangles with scaling points can be solved in expected O(n) time by a randomized algorithm. The 3-center problem as an LP-type problem. Given a nonempty 3-pierceable set R of axis-parallel rectangles, let v R be the rst point v in the sequence (v RT ; v RB ; v LT ; v LB ) such that the set of rectangles in R disjoint from v, denoted by R v, is 2-pierceable (v R exists; see Subsection 2.1). We put R = i, if v R's occurrence in this sequence is in position i, for 1 i 4. We extend this notion to ; = 0, v ; =?, and for R not 3-pierceable, R = 5, v R =?. Let R be a set of axis-parallel rectangles with scaling points. For G R, we dene u(g) = ( 000 G ; x L ; x R ; y B ; y T ; G 000 ; ~w(g 000 v G 000)); where G 000 = G( 000 G ), where x L, x R, y B, and y T are as dened above, for the scaled set G 000, and where ~w is the tuple w without its rst component (here there is no need to specify another scaling factor). As before, one can show that u(g) describes a unique candidate piercing set. Ordering on tuples (; x 1; x 2; y 1; y 2; ; ~w) is again lexicographic from left to right. Monotonicity follows easily, as in the 2-center case. Locality follows again, because u(g) encodes an explicit piercing set which determines whether u(g [ frg) > u(g) for some r (this happens i r( 000 G ) is not pierced by that set). For the combinatorial dimension, we have to nd a bound on the number of rectangles in a set G necessary for determining u(g): 16 rectangles enforce 000 G ; 4 rectangles x the lines `X; for each = 2; 3; 4, we need up to 5 rectangles for each preceding vertex v XY, to show that the rectangles disjoint from v XY are not 2-pierceable, thus enforcing the value of. Finally, we need 4+4 more rectangles to enforce the value of ~w (see the previous analysis). This gives a bound of = 43 (which, again, may be improved by a more careful analysis). It requires an actual implementation to decide whether the randomized algorithm derived from the LP-type framework does indeed have an impractical constant of proportionality in the linear time bound (we expect that this is not the case). At this point we have the following theoretical result. Theorem 3.7 The general rectilinear 3-center problem for n rectangles with scaling points can be solved in expected O(n) time by a randomized algorithm. The algorithms in Theorems 3.6 and 3.7 can be derandomized at the cost of an increase in the constant of proportionality only; this is done using the technique of [5]. (Note that this technique requires some extra condition beyond the LP-type axioms: For each basis B, let G B be the set of all constraints not violated by B. Then we require that the set system fg Bg B have nite VC-dimension. It is easily veried that this condition holds for the above problems.) Remark. One can apply Lemma 3.2(c), using a machinery similar to that above, to obtain a linear-time randomized algorithm for the general rectilinear 2-center problem in any dimension d. See also a related piercing result in [21]. 3.2 Rectilinear 4- and 5-center problems Let S be a set of n points in the plane. We want to nd the smallest, so that S can be covered by the union of four (or ve) axis-parallel squares of side length. We solve this problem by providing a general scheme for transforming a p-piercing algorithm for translates of squares to a rectilinear p-center algorithm. It is based on the matrix-searching technique of Frederickson and Johnson [13, 14, 15] (a previous recent application of that technique can be found in [17]). Specically, we observe that if S can be covered by the union of p squares of size, then it can also be covered by the union of p squares of size smaller than, unless is equal to the dierence between either the x-coordinates or the y-coordinates of a pair of points in S. 8

9 We therefore consider two nn matrices X and Y, dened as follows. Let (x 1; : : : ; x n) be the sequence of x-coordinates of the points of S, sorted in increasing order. Then X ij = x j x i if j > i and X ij = 0 otherwise. The matrix Y is dened symmetrically by the sorted y-coordinates of the points of S. We note that both matrices are monotone, in the sense that each row (resp. each column) is a monotone nondecreasing (resp. nonincreasing) sequence, and that the optimal is an entry of one of these matrices. Moreover, comparing with another can be done by applying a p-piercing algorithm to the set Q() = fs + Q j s 2 Sg, where Q is an axis-parallel square of size 1, centered at the origin. That is, Q() is p-pierceable if and only if. We can therefore apply the matrix searching technique of Frederickson and Johnson, as described in [13, 14, 15]. It nds the optimal by making only O(log n) calls to the p-piercing decision procedure. That is, we have shown: Theorem 3.8 If the p-piercing problem for n translates of a square can be solved in time t p(n), then the rectilinear p- center problem can be solved in O(t p(n) log n) time. Based on Theorems 2.3(b) and 2.5, we get (the proof of the lower bound will be given in the full version of the paper): Corollary 3.9 The rectilinear 4-center problem can be solved in time O(n log n), which is worst-case optimal, and the rectilinear 5-center problem can be solved in time O(n log 5 n). Remark. The Frederickson-Johnson technique fails in the weighted case, where we need to apply full-edged parametric searching [29] (see also [1, 2]). This results in O(n polylog n) solutions for both the weighted (or general) rectilinear 4- and 5-center problems. Details are given in the full version. 4 Polygonal 2-Piercing and 2- Center Problems In this section we consider the 2-piercing and 2-center problems for convex polygons. In the 2-center problem, we are given a set S of n points in the plane, and a convex c-gon P, and we want to cover S with two homothetic copies of P whose maximum size is as small as possible. An equivalent formulation of the problem is as follows. Let P ~ be the reected image of P through the origin. We want to nd the smallest scaling factor for which the collection P() = fs+ P ~ j s 2 Sg of isothetic polygons is 2-pierceable. As a matter of fact, we consider the following more general 2-piercing problem: A collection Q of convex polygons is called c-oriented if the orientations of the edges of each Q 2 Q belong to a xed set of c orientations. (We will also allow degenerate cases, in which some members of Q may have fewer than c edges; we treat these cases by regarding each member of Q as a convex c-gon, where some of its sides have zero length.) Let Q = fq 1; : : : ; Q ng be a collection of c-oriented convex polygons. We want to determine whether Q is 2-pierceable. Unlike the case where the given orientations come from the Figure 3: The arrangement A(L) for a 2-pierceable set of 4-oriented quadrilaterals sides of a square (which is the rectilinear 2-piercing problem treated earlier), there is no general Helly-type result for 2-piercing of Q (such a result was conjectured by Danzer and Grunbaum [6] and recently refuted by Katchalski and Nashtir [20]). However, such a Helly-type result was established in [20] for the case of homothetic triangles: if any 9 (or fewer) triangles in Q are 2-pierceable then Q is 2-pierceable. This result implies that the 2-center problem for homothets of some triangle can be solved in expected linear time, using the machinery of LP-type problems. However, our analysis will imply a simpler (and deterministic) linear-time solution for triangles. For each edge orientation, let L be the directed line at orientation that satises the following properties: L passes through an edge e of some polygon Q 2 Q (including degenerate edges, as above), so that Q lies to the left of L. No edge at orientation of any polygon in Q is contained in the open halfplane to the left of L. Let L be the resulting collection of c lines, and let A(L) denote their arrangement. See Figure 3. Note that if Q can be pierced by two points a, b, then the closed left halfplane bounded by each of the lines L must contain either a or b (or else the corresponding polygon Q will not be pierced). Lemma 4.1 If Q can be pierced by two points, then it can be pierced by a pair of points lying on the lines of L. For Q a collection of homothetic triangles or 4-oriented trapezoids, we may assume that one of the piercing points lies at a vertex of A(L). Proof: Let a and b be two points that pierce Q. We will show that a can be moved to an appropriate edge g of A(L), on the boundary of the face containing a, so that, during this motion, we do not exit any polygon in Q. A similar property also holds for b, and this clearly implies the assertion of the lemma. The following claim is immediate from the denitions: Claim: (i) Suppose that a lies in the left halfplane of some line L. If a moves arbitrarily within this halfplane, it cannot exit any polygon of Q through an edge at orientation. (ii) Suppose that a lies in the right halfplane of some line L. 9